...
- fD = 2fMAX
- fD = fMAX
- fD = 4fMAX
Riešenie
1a)
1b)
Maximálna frekvencia prítomná v signály: fM = 3 Hz
...
Vzorkovanie signálu: f(t) →f(nTD ) = cos(2π.nTD )+1/4 cos(2π.3nTD)
1c)
fD = 2fMAX = 2.3Hz = 6Hz
TD = 1/6 s
n | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
nTd | 0 | 0.167 | 0.333 | 0.5 | 0.667 | 0.833 | 1 | 1.167 | 1.333 | 1.5 | 1.667 | 1.833 | 1.167 |
f(nTd) | 1.25 | 0.25 | -0.25 | -1.25 | -0.25 | 0.25 | 1.25 | 0.25 | -0.25 | -1.25 | -0.25 | -1.25 | -0.25 |
fD = fMAX= 3Hz
TD = 1/3 s
n | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
nTd | 0 | 0.333 | 0.667 | 1 | 1.333 | 1.667 | 2 | 2.333 | 2.667 | 3 | 3.333 | 3.667 | 4 |
f(nTd) | 1.25 | -0.25 | -0.25 | 1.25 | -0.25 | -0.25 | 1.25 | -0.25 | -0.25 | 1.25 | -0.25 | -0.25 | 1.25 |
fD = 4fMAX = 4.3Hz = 12Hz
TD = 1/12 s
n | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
nTd | 0 | 0.083 | 0.167 | 0.25 | 0.333 | 0.417 | 0.5 | 0.583 | 0.667 | 0.75 | 0.833 | 0,917 | 1 |
f(nTd) | 1.25 | -0.25 | -0.25 | 1.25 | -0.25 | -0.25 | 1.25 | -0.25 | -0.25 | 1.25 | -0.25 | -0.25 | 1.25 |
Úloha 2
Vykonajte rekonštrukciu vzoriek pre signál f(t) z úlohy 1c) pomocou Shannon-Kotelníkového rádu.
Shannon-Kotelnikov rád:
fD = 2fMAX
DOMÁCA ÚLOHA - SK rád
fD = fMAX
fD =4fMAX
Úloha 3
Pre signál f(t) vykonajte úlohy:
A. Zakreslenie signálu.
B. Odvodenie minimálnej vhodnej vzorkovacej frekvencie.
C. Hodnoty funkcie pre vzorkovacie frekvencie pre n ∈⟨0,12⟩:
D. Rekonštrukcia pomocou Shannon-Kotelnikového rádu.
Riešenie
3a)
3b)
Maximálna frekvencia prítomná v signály: fM = 4 Hz
Minimálna vzorkovacia frekvencia: fD ≥ 2fM → fD ≥ 8 Hz
Maximálna vzorkovacia perióda: TD ≤1/fD → TD ≤1/8
Vzorkovanie signálu: f(t) →f(nTD ) = cos(2π.2nTD )+1/2 cos(2π.4nTD)
3c)
fD = 2fMAX
n | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
nTd |
f(nTd)
1
| 0 |
| 0.125 |
| 0.25 |
fD = fMAX
n
0
1
2
3
4
5
6
7
8
9
10
11
12
nTd
0
0.333
| 0.375 | 0.5 | 0.625 | 0.75 | 0.875 | 1 | 1. |
| 125 | 1. |
| 25 |
2
| 1. |
| 375 |
2.667
3
3.333
3.667
4
| 1.5 | |
f(nTd) | 1 |
| 0 |
| -1 | 0 |
| 1 |
| 0 |
| -1 | 0 |
| 1 |
| 0 |
| - |
| 1 |
| 0 |
| 1 |
...
fD = 4fMAX
n | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
nTd | 0 | 0. |
0625 | 0. |
125 | 1.1875 | 0.25 | 0. |
3125 | 0. |
375 | 0. |
4375 | 0. |
5 | 0. |
5625 | 0. |
625 | 0. |
6815 | 0 |
.75 |
1
f(nTd) | 1 | 1. |
207 |
0 |
- |
1. |
207 | -1 |
-0. |
207 | 0 |
0. |
207 | 1 | 1. |
207 |
0 |
- |
1. |
1.25
-0.25
-0.25
1.25
...
207 |
3d)
fD = 2fMAX
fD = 4fMAX
